A small solid sphere of mass M0, of radius R0, and of uniform density ?0 is plac

Question

A small solid sphere of mass M0, of radius R0, and of uniform density ?0 is placed in a large bowl containing water. It floats and the level of the water in the dish is L. Given the information below, determine the possible effects on the water level L, (R-Rises, F-Falls, U-Unchanged), when that sphere is replaced by a new solid sphere of uniform density.


The new sphere has radius R = R0 and mass M > M0
The new sphere has mass M > M0 and density ? = ?0
The new sphere has density ? > ?0 and radius R < R0
The new sphere has mass M = M0 and density ? > ?0
The new sphere has density ? < ?0 and mass M = M0
The new sphere has radius R < R0 and density ? = ?0

Explanation / Answer

The density of an object determines the amount of the object which is under water. Ratio of density of object to density of water = fraction of volume of object which is under water. When more of the object is under water, the level of the water in the bowl is higher! When less of the object is under water, the level of the water in the bowl is lower! #1 Increased density ? decreases fraction of volume of object which is under water ? water level rises. Density = mass ÷ Volume Volume of Sphere = 4/3 * p * r^3 #2. As radius increases, the volume increases, the density decreases, fraction under decreases, so water level lowers. #3. As mass increases, the density increases, fraction under increases, so water level rises The new sphere has density ? < ?0 and radius R > R0 Decrease of density = up, increase of radius = up

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